# Drawing Sorgenfrey continuous functions

The Sorgenfrey line is a well known topological space. It is the real number line with open intervals defined as sets of the form $[a,b)$. Though this is a seemingly small tweak, it generates a vastly different space than the usual real number line. In this post, we look at the Sorgenfrey line from the continuous function perspective, in particular, the continuous functions that map the Sorgenfrey line into the real number line. In the process, we obtain insight into the space of continuous functions on the Sorgenfrey line.

The next post is a continuation on the theme of drawing Sorgenfrey continuous functions.

The Sorgenfrey Line

Let $\mathbb{R}$ denote the real number line. The usual open intervals are of the form $(a,b)=\left\{x \in \mathbb{R}: a. The union of such open intervals is called an open set. If more than one topologies are considered on the real line, these open sets are referred to as the usual open sets or Euclidean open sets (on the real line). The open intervals $(a,b)$ form a base for the usual topology on the real line. One important fact abut the usual open sets is that the usual open sets can be generated by the intervals $(a,b)$ where both end points are rational numbers. Thus the usual topology on the real line is said to have a countable base.

Now tweak the usual topology by calling sets of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x open intervals. Then form open sets by taking unions of all such open intervals. The collection of such open sets is called the Sorgenfrey topology (on the real line). The real number line $\mathbb{R}$ with the Sorgenfry topology is called the Sorgenfrey line, denoted by $\mathbb{S}$. The Sorgenfrey line has been discussed in this blog, starting with this post. This post examines continuous functions from $\mathbb{S}$ into the real line. In the process, we gain insight on the space of continuous functions defined on $\mathbb{S}$.

Note that any usual open interval $(a,b)$ is the union of intervals of the form $[c,d)$. Thus any usual (Euclidean) open set is an open set in the Sorgenfrey line. Thus the usual topology (on the real line) is contained in the Sorgenfrey topology, i.e. the usual topology is a weaker (coarser) topology.

Let $C(\mathbb{R})$ be the set of all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}$ where the domain is the real number line with the usual topology. Let $C(\mathbb{S})$ be the set of all continuous functions $f:\mathbb{S} \rightarrow \mathbb{R}$ where the domain is the Sorgenfrey line. In both cases, the range is always the number line with the usual topology. Based on the preceding paragraph, any continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is also continuous with respect to the Sorgenfrey line, i.e. $C(\mathbb{R}) \subset C(\mathbb{S})$.

Pictures of Continuous Functions

Consider the following two continuous functions.

Figure 1 – CDF of the standard normal distribution

Figure 2 – CDF of the uniform distribution

The first one (Figure 1) is the cumulative distribution function (CDF) of the standard normal distribution. The second one (Figure 2) is the CDF of the uniform distribution on the interval $(0,a)$ where $a>0$. Both of these are continuous in the usual Euclidean topology (in the domain). Such graphs would make regular appearance in a course on probability and statistics. They also show up in a calculus course as an everywhere differentiable curve (Figure 1) and as a differentiable curve except at finitely many points (Figure 2). Both of these functions can also be regarded as continuous functions on the Sorgenfrey line.

Consider a function that is continuous in the Sorgenfrey line but not continuous in the usual topology.

Figure 3 – Right continuous function

Figure 3 is a function that maps the interval $(-\infty,0)$ to -1 and maps the interval $[0,\infty)$ to 1. It is not continuous in the usual topology because of the jump at $x=0$. But it is a continuous function when the domain is considered to be the Sorgenfrey line. Because of the open intervals being $[a,b)$, continuous functions defined on the Sorgenfrey line are right continuous.

The cumulative distribution function of a discrete probability distribution is always right continuous, hence continuous in the Sorgenfrey line. Here’s an example.

Figure 4 – CDF of a discrete uniform distribution

Figure 4 is the CDF of the uniform distribution on the finite set $\left\{0,1,2,3,4 \right\}$, where each point has probability 0.2. There is a jump of height 0.2 at each of the points from 0 to 4. Figure 3 and Figure 4 are step functions. As long as the left point of a step is solid and the right point is hollow, the step functions are continuous on the Sorgenfrey line.

The take away from the last four figures is that the real-valued continuous functions defined on the Sorgenfrey line are right continuous and that step functions (with the left point solid and the right point hollow) are Sorgenfrey continuous.

A Family of Sorgenfrey Continuous Functions

The four examples of continuous functions shown above are excellent examples to illustrate the Sorgenfrey topology. We now introduce a family of continuous functions $f_a:\mathbb{S} \rightarrow \mathbb{R}$ for $0. These continuous functions will lead to additional insight on the function space whose domain space is the Sorgenfrey line.

For any $0, the following gives the definition and the graph of the function $f_a$.

$\displaystyle f_a(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ -\infty

Figure 5 – a family of Sorgenfrey continuous functions

Function Space on the Sorgenfrey Line

This is the place where we switch the focus to function space. The set $C(\mathbb{S})$ is a subset of the product space $\mathbb{R}^\mathbb{R}$. So we can consider $C(\mathbb{S})$ as a topological space endowed with the topology inherited as a subspace of $\mathbb{R}^\mathbb{R}$. This topology on $C(\mathbb{S})$ is called the pointwise convergence topology and $C(\mathbb{S})$ with the product subspace topology is denoted by $C_p(\mathbb{S})$. See here for comments on how to work with the pointwise convergence topology.

For the present discussion, all we need is some notation on a base for $C_p(\mathbb{S})$. For $x \in \mathbb{S}$, and for any open interval $(a,b)$ (open in the usual topology of the real number line), let $[x,(a,b)]=\left\{h \in C_p(\mathbb{S}): h(x) \in (a, b) \right\}$. Then the collection of intersections of finitely many $[x,(a,b)]$ would form a base for $C_p(\mathbb{S})$.

The following is the main fact we wish to establish.

The function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum. In particular, the set $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$.

The above result will derive several facts on the function space $C_p(\mathbb{S})$, which are discussed in a section below. More interestingly, the proof of the fact that $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$ is based purely on the definition of the functions $f_a$ and the Sorgenfrey topology. The proof given below does not use any deep or high powered results from function space theory. So it should be a nice exercise on the Sorgenfrey topology.

I invite readers to either verify the fact independently of the proof given here or follow the proof closely. Lots of drawing of the functions $f_a$ on paper will be helpful in going over the proof. In this one instance at least, drawing continuous functions can help gain insight on function spaces.

Working out the Proof

The following diagram was helpful to me as I worked out the different cases in showing the discreteness of the family $F=\left\{f_a: 0. The diagram is a valuable aid in convincing myself that a given case is correct.

Figure 6 – A comparison of three Sorgenfrey continuous functions

Now the proof. First, $F$ is relatively discrete in $C_p(\mathbb{S})$. We show that for each $a$, there is an open set $O$ containing $f_a$ such that $O$ does not contain $f_w$ for any $w \ne a$. To this end, let $O=[a,V_1] \cap [-a,V_2]$ where $V_1$ and $V_2$ are the open intervals $V_1=(0.9,1.1)$ and $V_2=(-0.1,0.1)$. With Figure 6 as an aid, it follows that for $0, $f_b \notin O$ and for $a, $f_c \notin O$.

The open set $O=[a,V_1] \cap [-a,V_2]$ contains $f_a$, the function in the middle of Figure 6. Note that for $0, $f_b(-a)=1$ and $f_b(-a) \notin V_2$. Thus $f_b \notin O$. On the other hand, for $a, $f_c(a)=0$ and $f_c(a) \notin V_1$. Thus $f_c \notin O$. This proves that the set $F$ is a discrete subspace of $C_p(\mathbb{S})$ relative to $F$ itself.

Now we show that $F$ is closed in $C_p(\mathbb{S})$. To this end, we show that

for each $g \in C_p(\mathbb{S})$, there is an open set $U$ containing $g$ such that $U$ contains at most one point of $F$.

Actually, this has already been done above with points $g$ that are in $F$. One thing to point out is that the range of $f_a$ is $\left\{0,1 \right\}$. As we consider $g \in C_p(\mathbb{S})$, we only need to consider $g$ that maps into $\left\{0,1 \right\}$. Let $g \in C_p(\mathbb{S})$. The argument is given in two cases regarding the function $g$.

Case 1. There exists some $a \in (0,1)$ such that $g(a) \ne g(-a)$.

We assume that $g(a)=0$ and $g(-a)=1$. Then for all $0, $f_b(a)=1$ and for all $a, $f_c(-a)=0$. Let $U=[a,(-0.1,0.1)] \cap [-a,(0.9,1.1)]$. Then $g \in U$ and $U$ contains no $f_b$ for any $0 and $f_c$ for any $a. To help see this argument, use Figure 6 as a guide. The case that $g(a)=1$ and $g(-a)=0$ has a similar argument.

Case 2. For every $a \in (0,1)$, we have $g(a) = g(-a)$.

Claim. The function $g$ is constant on the interval $(-1,1)$. Suppose not. Let $0 such that $g(a) \ne g(b)$. Suppose that $0=g(b) < g(a)=1$. Consider $W=\left\{w. Clearly the number $a$ is an upper bound of $W$. Let $u \le a$ be a least upper bound of $W$. The function $g$ has value 1 on the interval $(u,a)$. Otherwise, $u$ would not be the least upper bound of the set $W$. There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=0$ for all $n$. Otherwise, $u$ would not be the least upper bound of the set $W$.

It follows that $g(u)=1$. Otherwise, the function $g$ is not continuous at $u$. Now consider the 6 points $-a<-u<-b. By the assumption in Case 2, $g(u)=g(-u)=1$ and $g(b)=g(-b)=0$. Since $g(x_n)=0$ for all $n$, $g(-x_n)=0$ for all $n$. Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=0$, contradicting $g(-u)=1$. Thus we cannot have $0=g(b) < g(a)=1$.

Now suppose we have $1=g(b) > g(a)=0$ where $0. Consider $W=\left\{w. Clearly $W$ has an upper bound, namely the number $a$. Let $u \le a$ be a least upper bound of $W$. The function $g$ has value 0 on the interval $(u,a)$. Otherwise, $u$ would not be the least upper bound of the set $W$. There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=1$ for all $n$. Otherwise, $u$ would not be the least upper bound of the set $W$.

It follows that $g(u)=0$. Otherwise, the function $g$ is not continuous at $u$. Now consider the 6 points $-a<-u<-b. By the assumption in Case 2, $g(u)=g(-u)=0$ and $g(b)=g(-b)=1$. Since $g(x_n)=1$ for all $n$, $g(-x_n)=1$ for all $n$. Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=1$, contradicting $g(-u)=0$. Thus we cannot have $1=g(b) > g(a)=0$.

The claim that the function $g$ is constant on the interval $(-1,1)$ is established. To wrap up, first assume that the function $g$ is 1 on the interval $(-1,1)$. Let $U=[0,(0.9,1.1)]$. It is clear that $g \in U$. It is also clear from Figure 5 that $U$ contains no $f_a$. Now assume that the function $g$ is 0 on the interval $(-1,1)$. Since $g$ is Sorgenfrey continuous, it follows that $g(-1)=0$. Let $U=[-1,(-0.1,0.1)]$. It is clear that $g \in U$. It is also clear from Figure 5 that $U$ contains no $f_a$.

We have established that the set $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$.

What does it Mean?

The above argument shows that the set $F$ is a closed an discrete subspace of the function space $C_p(\mathbb{S})$. We have the following three facts.

 Three Results $C_p(\mathbb{S})$ is separable. $C_p(\mathbb{S})$ is not hereditarily separable. $C_p(\mathbb{S})$ is not a normal space.

To show that $C_p(\mathbb{S})$ is separable, let’s look at one basic helpful fact on $C_p(X)$. If $X$ is a separable metric space, e.g. $X=\mathbb{R}$, then $C_p(X)$ has quite a few nice properties (discussed here). One is that $C_p(X)$ is hereditarily separable. Thus $C_p(\mathbb{R})$, the space of real-valued continuous functions defined on the number line with the pointwise convergence topology, is hereditarily separable and thus separable. Recall that continuous functions in $C_p(\mathbb{R})$ are also Soregenfrey line continuous. Thus $C_p(\mathbb{R})$ is a subspace of $C_p(\mathbb{S})$. The space $C_p(\mathbb{R})$ is also a dense subspace of $C_p(\mathbb{S})$. Thus the space $C_p(\mathbb{S})$ contains a dense separable subspace. It means that $C_p(\mathbb{S})$ is separable.

Secondly, $C_p(\mathbb{S})$ is not hereditarily separable since the subspace $F=\left\{f_a: 0 is a closed and discrete subspace.

Thirdly, $C_p(\mathbb{S})$ is not a normal space. According to Jones’ lemma, any separable space with a closed and discrete subspace of cardinality of continuum is not a normal space (see Corollary 1 here). The subspace $F=\left\{f_a: 0 is a closed and discrete subspace of the separable space $C_p(\mathbb{S})$. Thus $C_p(\mathbb{S})$ is not normal.

Remarks

The topology of the Sorgenfrey line is vastly different from the usual topology on the real line even though the the Sorgenfrey topology is obtained by a seemingly small tweak from the usual topology. The real line is a metric space while the Sorgenfrey line is not metrizable. The real number line is connected while the Sorgenfrey line is not. The countable power of the real number line is a metric space and thus a normal space. On the other hand, the Sorgenfrey line is a classic example of a normal space whose square is not normal. See here for a basic discussion of the Sorgenfrey line.

The pictures of Sorgenfrey continuous functions demonstrated here show that the real number line and the Sorgenfrey line are also very different from a function space perspective. The function space $C_p(\mathbb{R})$ has a whole host of nice properties: normal, Lindelof (hence paracompact and collectionwise normal), hereditarily Lindelof (hence hereditarily normal), hereditarily separable, and perfectly normal (discussed here).

Though separable, the function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum, making it not hereditarily separable and not normal.

For more information about $C_p(X)$ in general and $C_p(\mathbb{S})$ in particular, see [1] and [2]. A different proof that $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum can be found in Problem 165 in [2].

The next post is a continuation on the theme of drawing Sorgenfrey continuous functions.

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

# A strategy for finding CCC and non-separable spaces

In this post we present a general strategy for finding CCC spaces that are not separable. As illustration, we give four implementations of this strategy.

In searching for counterexamples in topology, one good place to look is of course the book by Steen and Seebach [2]. There are four examples of spaces that are CCC but not separable found in [2] – counterexamples 20, 21, 24 and 63. Counterexamples 20 and 21 are not Hausdorff. Counterexample 24 is the uncountable Fort space (it is completely normal but not perfectly normal). Counterexample 63 (Countable Complement Extension Topology) is Hausdorff but is not regular. These are valuable examples especially the last two (24 and 63). The examples discussed below expand the offerings in Steen and Seebach.

The discussion of CCC but not separable in this post does not use axioms beyond the usual axioms of set theory (i.e. ZFC). The discussion here does not touch on Suslin lines or other examples that require extra set theory. The existence of Suslin lines is independent of ZFC. A Suslin line would produce an example of a perfectly normal first countable CCC non-separable space. In models of set theory where Suslin lines do not exist, a perfectly normal first countable CCC non-separable space can also be produced using other set-theoretic assumptions. The examples discussed below are not as nice as the set-theoretic examples since they usually are not first countable and perfectly normal.

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The countable chain conditon

A topological space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of discussion, if $X$ has the CCC, we also say that $X$ is a CCC space or X is CCC. A space $X$ is separable if there exists a countable subset $A$ of $X$ such that $A$ is dense in $X$ (meaning that if $U$ is a nonempty open subset of $X$, then $U \cap A \ne \varnothing$).

It is clear that any separable space has the CCC. In metric spaces, these two properties are equivalent. Among topological spaces in general, the two properties are not identical. Thus “CCC but not separable” is one way to distinguish between metrizable spaces and non-metrizable spaces. Even in non-metrizable spaces, “CCC but not separable” is also a way to obtain more information about the spaces being investigated.

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The strategy

Here’s the strategy for finding CCC and not separable.

The strategy is to narrow the focus to spaces where “CCC and not separable” is likely to exist. Specifically, look for a space or a class of spaces such that each space in the class has the countable chain condition but is not hereditarily separable. If the non-separable subspace is also a dense subspace of the starting space, it would be “CCC and not separable.”

Any dense subspace of a CCC space always has the CCC. Thus the search focuses on the subspaces in a CCC space that are reliably CCC. The strategy is to find non-separable spaces among these dense subspaces. The search is given an assist if the space or class of spaces in question has a characteristic that delineate the “separable” from the CCC (see Example 3 and Example 4 below).

In the following sections, we illustrate four different ways to apply the strategy.

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Example 1

The first way is a standard example found in the literature. The space to start from is the product space of separable spaces, which is always CCC. By a theorem of Ross and Stone, the product of more than continuum many separable spaces is not separable. Thus one way to get an example of CCC but not separable space is to take the product of more than continuum many separable spaces. For example, if $c$ is the cardinality of continuum, then consider $\left\{0,1 \right\}^{2^c}$, the product of $2^c$ many copies of $\left\{0,1 \right\}$, or consider $X^{2^c}$ where $X$ is your favorite separable space.

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Example 2

The second implementation of the strategy is also from taking the product of separable spaces. This time the number of factors does not have to be more than continuum. Here, we focus on one particular dense subspace of the product space, the $\Sigma$-products. To make this clear, let’s focus on a specific example. Consider $X=\left\{0,1 \right\}^{c}$ where $c$ is the cardinality of continuum. Consider the following subspace.

$\Sigma(\left\{0,1 \right\}^{c})= \left\{x \in X: x(\alpha) \ne 0 \text{ for at most countably many } \alpha < c \right\}$

The subspace $\Sigma(\left\{0,1 \right\}^{c})$ is dense in $X$, thus has CCC. It is straightforward to verify that $\Sigma(\left\{0,1 \right\}^{c})$ is not separable.

To implement this example, find any space $X$ which is a product space of separable spaces, each of which has at least two point (one of the points is labeled 0). The dense subspace is the $\Sigma$-product, which is the subspace consisting of all points that are non-zero at only countably many coordinates. The $\Sigma$-product has the countable chain condition since it is a dense subspace of the CCC space $X$. The $\Sigma$-product is not separable since there are uncountably many factors in the product space $X$ and that each factor has at least two points. This idea had been implemented in this previous post.

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Example 3

The third class of spaces is the class of Pixley-Roy spaces, which are hyperspaces. Given a space $X$, let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For $F \in \mathcal{F}[X]$ and for any open subset $U$ of $X$, let $[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$. The sets $[F,U]$ over all $F$ and $U$ form a base for a topology on $\mathcal{F}[X]$. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space.

The Pixley-Roy hyperspaces are discussed in this previous post. Whenever the ground space $X$ is uncountable, $\mathcal{F}[X]$ is not a separable space. We need to identify the $\mathcal{F}[X]$ that are CCC. According to the previous post on Pixley-Roy hyperspaces, for any space $X$ with a countable network, $\mathcal{F}[X]$ is CCC. Thus for any uncountable space $X$ with a countable network, the Pixley-Roy space $\mathcal{F}[X]$ is a CCC space that is not separable. The following gives a few such examples.

$\mathcal{F}[\mathbb{R}]$

$\mathcal{F}[X]$ where $X$ is any uncountable, separable and metrizable space.

$\mathcal{F}[X]$ where $X$ is uncountable and is the continuous image of a separable metrizable space.

Spaces with countable networks are discussed in this previous post. An example of a space $X$ that is the continuous image of a separable metrizable space is the bow-tie space found this previous post. Another example is any quote space of a separable metrizable space.

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Example 4

For the fourth implementation of the strategy, we go back to the product space of separable spaces in Example 2, with the exception that the focus is on the product of the real line $\mathbb{R}$. Let $X$ be any uncountable completely regular space. The product space $\mathbb{R}^X$ always has the CCC since it is a product of separable space. Now we single out a dense subspace of $\mathbb{R}^X$ for which there is a characterization for separability, namely the subspace $C(X)$, which is the set of all continuous functions from $X$ into $\mathbb{R}$. The subspace $C(X)$ as a topological space is usually denoted by $C_p(X)$. For a basic discussion of $C_p(X)$, see this previous post.

We know precisely when $C_p(X)$ is separable. The following theorem captures the idea.

Theorem 1 – Theorem I.1.3 [1]
The function space $C_p(X)$ is separable if and only if the domain space $X$ has a weaker (or coarser) separable metric topology (in other words, $X$ is submetrizable with a separable metric topology).

Based on the theorem, $C_p(X)$ is separable for any separable metric space $X$. Other examples of separable $C_p(X)$ include spaces $X$ that are created by tweaking the usual Euclidean topology on the real line and at the same time retaining the usual real line topology as a weaker topology, e.g. the Sorgenfrey line and the Michael line. Thus $C_p(X)$ would be separable if $X$ is a space such as the Sorgenfrey line or the Michael line. For our purpose at hand, we need to look for spaces that are not like the Sorgenfrey line or the Michael line. Here’s some examples of spaces $X$ that have no weaker separable metric topology.

• Any compact space $X$ that is not metrizable.
• The space $X=\omega_1$, the first uncountable ordinal with the order topology.
• Any space $X=C_p(Y)$ where $Y$ is not separable.

The function space $C_p(X)$ for any one of the above three spaces has the CCC but is not separable. It is well known that any compact space with a weaker metrizable topology is metrizable. Some examples for compact $X$ are: the first uncountable successor ordinal $\omega_1+1$, the double arrow space, and the product space $\left\{0,1 \right\}^{\omega_1}$.

It can be shown that $C_p(\omega_1)$ is not separable (see this previous post). The last example is due to the following theorem.

Theorem 2 – Theorem I.1.4 [1]
The function space $C_p(Y)$ has a weaker (or coarser) separable metric topology if and only if the domain space $Y$ is separable.

Thus picking a non-separable space $Y$ would guarantee that $C_p(Y)$ has no weaker separable metric topology. As a result, $C_p(C_p(Y))$ is a CCC and not separable space.

Interestingly, Theorem 1 and Theorem 2 show a duality existing between the property of having a weaker separable metric topology and the property of being separable. The two theorems allow us to switch the two properties between the domain space and the function space.

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Remarks

Another interesting point to make is that Theorem 1 and Theorem 2 together allow us to “buy one get one free.” Once we obtain a space $X$ that is CCC and not separable from any one of the avenues discussed here, the function space $C_p(X)$ has no weaker separable metric topology (by Theorem 2) and the function space $C_p(C_p(X))$ is another example of CCC and not separable.

The strategy discussed above unifies all four examples. Undoubtedly there will be other examples that can come from the strategy. To find more examples, find a space or a class of spaces that are reliably CCC and then look for potential non-separable spaces among the dense subspaces of the starting space.

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Exercises

1. Show that in metrizable spaces, CCC and separable are equivalent. The only part that needs to be shown is that if $X$ is metrizable and CCC, then $X$ is separable.
2. Show that any dense subspace of a CCC space is also CCC.
3. Verify that the space $\Sigma(\left\{0,1 \right\}^{c})$ defined in Example 2 is dense in $X$ and is not separable.
4. Verify that the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ defined in Example 3 is CCC and not separable.
5. Verify that function space $C_p(\omega_1)$ mentioned in Example 4 is not separable. Hint: use the pressing down lemma.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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$\copyright \ 2016 \text{ by Dan Ma}$
Revised October 26, 2018

# Cp(X) where X is a separable metric space

Let $\tau$ be an uncountable cardinal. Let $\prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau}$ be the Cartesian product of $\tau$ many copies of the real line. This product space is not normal since it contains $\prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1}$ as a closed subspace. However, there are dense subspaces of $\mathbb{R}^{\tau}$ are normal. For example, the $\Sigma$-product of $\tau$ copies of the real line is normal, i.e., the subspace of $\mathbb{R}^{\tau}$ consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of $\mathbb{R}^{\tau}$ that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space $C_p(X)$ where $X$ is a separable metrizable space.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

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$C_p(X)$ when $X$ is a separable metric space

In the remainder of the post, $X$ denotes a separable metrizable space. Then, $C_p(X)$ is more than normal. The function space $C_p(X)$ has the following properties:

• normal,
• Lindelof (hence paracompact and collectionwise normal),
• hereditarily Lindelof (hence hereditarily normal),
• hereditarily separable,
• perfectly normal.

All such properties stem from the fact that $C_p(X)$ has a countable network whenever $X$ is a separable metrizable space.

Let $L$ be a topological space. A collection $\mathcal{N}$ of subsets of $L$ is said to be a network for $L$ if for each $x \in L$ and for each open $O \subset L$ with $x \in O$, there exists some $A \in \mathcal{N}$ such that $x \in A \subset O$. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for $C_p(X)$, let $\mathcal{B}$ be a countable base for the domain space $X$. For each $B \subset \mathcal{B}$ and for any open interval $(a,b)$ in the real line with rational endpoints, consider the following set:

$[B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}$

There are only countably many sets of the form $[B,(a,b)]$. Let $\mathcal{N}$ be the collection of sets, each of which is the intersection of finitely many sets of the form $[B,(a,b)]$. Then $\mathcal{N}$ is a network for the function space $C_p(X)$. To see this, let $f \in O$ where $O=\bigcap_{x \in F} [x,O_x]$ is a basic open set in $C_p(X)$ where $F \subset X$ is finite and each $O_x$ is an open interval with rational endpoints. For each point $x \in F$, choose $B_x \in \mathcal{B}$ with $x \in B_x$ such that $f(B_x) \subset O_x$. Clearly $f \in \bigcap_{x \in F} \ [B_x,O_x]$. It follows that $\bigcap_{x \in F} \ [B_x,O_x] \subset O$.

Examples include $C_p(\mathbb{R})$, $C_p([0,1])$ and $C_p(\mathbb{R}^\omega)$. All three can be considered subspaces of the product space $\mathbb{R}^c$ where $c$ is the cardinality of the continuum. This is true for any separable metrizable $X$. Note that any separable metrizable $X$ can be embedded in the product space $\mathbb{R}^\omega$. The product space $\mathbb{R}^\omega$ has cardinality $c$. Thus the cardinality of any separable metrizable space $X$ is at most continuum. So $C_p(X)$ is the subspace of a product space of $\le$ continuum many copies of the real lines, hence can be regarded as a subspace of $\mathbb{R}^c$.

A space $L$ has countable extent if every closed and discrete subset of $L$ is countable. The $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ of the separable metric spaces $\left\{X_\alpha: \alpha \in A \right\}$ is a dense and normal subspace of the product space $\prod_{\alpha \in A} X_\alpha$. The normal space $\Sigma_{\alpha \in A} X_\alpha$ has countable extent (hence collectionwise normal). The examples of $C_p(X)$ discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Working with the function space Cp(X)

This post provides basic information about the space of real-valued continuous functions with the pointwise convergence topology. The goal is to discuss the setting and to define the standard basic open sets in the function space, providing background information for subsequent posts.

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Completely Regular Spaces

The starting point is a completely regular space. A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

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Defining the Function Space $C_p(X)$

Let $X$ be a completely regular space. Let $C(X)$ be the set of all real-valued continuous functions defined on the space $X$. The set $C(X)$ is naturally a subspace of the product space $\prod_{x \in X} \mathbb{R}=\mathbb{R}^X$. Thus $C(X)$ can be endowed with the subspace topology inherited from the product space $\mathbb{R}^X$. When this is the case, the function space $C(X)$ is denoted by $C_p(X)$. The topology on $C_p(X)$ is called the pointwise convergence topology.

Now we need a good handle on the open sets in the function space $C_p(X)$. A basic open set in the product space $\mathbb{R}^X$ is of the form $\prod_{x \in X} U_x$ where each $U_x$ is an open subset of $\mathbb{R}$ such that $U_x = \mathbb{R}$ for all but finitely many $x \in X$ (equivalently $U_x \ne \mathbb{R}$ for only finitely many $x \in X$). Thus a basic open set in $C_p(X)$ is of the form:

$C(X) \cap \biggl(\prod_{x \in X} U_x \biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where each $U_x$ is an open subset of $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x \in X$. In addition, when $U_x \ne \mathbb{R}$, we can take $U_x$ to be an open interval of the form $(a,b)$. To simplify notation, the basic open sets as described in (1) can also be notated by:

$\prod_{x \in X} U_x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1a)$

Thus when working with open sets in $C_p(X)$, we take $\prod_{x \in X} U_x$ to mean the set of all $f \in C(X)$ such that $f(x) \in U_x$ for each $x \in X$.

To make the basic open sets of $C_p(X)$ more explicit, (1) or (1a) is translated as follows:

$\bigcap_{x \in F} \ [x, O_x] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

where $F \subset X$ is a finite set, for each $x \in F$, $O_x$ is an open interval of $\mathbb{R}$, and $[x,O_x]$ is the set of all $f \in C(X)$ such that $f(x) \in O_x$.

There is another description of basic open sets that is useful. Let $f \in C_p(X)$. Let $F \subset X$ be finite. Let $\epsilon>0$. Let $B(f,F,\epsilon)$ be defined as follows:

$B(f,F,\epsilon)=\left\{g \in C(X): \forall \ x \in F, \lvert f(x)-g(x) \lvert< \epsilon \right\} \ \ \ \ \ \ \ (3)$

In proving results about $C_p(X)$, we can use basic open sets that are described in any one of the three forms (1), (2) and (3). If $U$ is a basic open subset of $C_p(X)$, as described in (1) or (1a), we use $supp(U)$ to denote the finite set of $x \in X$ such that $U_x \ne \mathbb{R}$. The set $supp(U)$ is called the support of $U$. The support for the basic open sets as described in (2) and (3) is already explicitly stated.

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Basic Discussion

The theory of $C_p(X)$ is a vast subject area. For a systematic introduction, see [1]. One fundamental theme in function space theory is the study of how properties of $X$ and $C_p(X)$ are related. The domain space $X$ and the function space $C(X)$ are not on the same footing. The domain $X$ only has a topological structure. The function space $C_p(X)$ carries a topology and two natural algebraic operations of addition and multiplication, making it a topological ring. In addition, $C_p(X)$ can be regarded as a topological group, or a linear topological space. In this post and in many subsequent posts, we narrow the focus to the topological properties of $X$ and $C_p(X)$, paying attention to the how the topological properties of $X$ and $C_p(X)$ are related.

In addition to the pointwise convergence topology, there are other topologies that can be defined on $C(X)$, e.g., the compact-open topology, the topology of uniform convergence and others. Both the pointwise convergence topology and the compact-open topology are examples of set-open topologies. In this post and in many of the subsequent posts, the focus is on the pointwise convergence topology, i.e., the subspace topology on $C(X)$ inherited from the product space.

The space $C_p(X)$ automatically inherits certain properties of the product space $\mathbb{R}^X$. Note that $C(X)$ is dense in $\mathbb{R}^X$. The product $\mathbb{R}^X$ has the countable chain condition (CCC) since it is a product of separable spaces. Hence $C_p(X)$ always has the CCC, i.e., there are no uncountably many pairwise disjoint open subsets of $C_p(X)$, regardless what the domain space $X$ is. One consequence of the CCC is that $C_p(X)$ is paracompact if and only if $C_p(X)$ is Lindelof.

It is well known that $\mathbb{R}^X$ is separable if and only if the cardinality of $X$ $\le$ continuum. Since $C(X)$ is dense in $\mathbb{R}^X$, $C_p(X)$ is not separable if the cardinality of $X$ $>$ continuum. Thus $C_p(X)$ is one way to get a CCC space that is not separable. There are non-separable $C_p(X)$ where the cardinality of $X$ $\le$ continuum. Obtaining such $C_p(X)$ would require more than the properties of the product space $\mathbb{R}^X$; using properties of $X$ would be necessary.

The properties of $C_p(X)$ discussed so far are inherited from the product space. Refer to chapter one of [1] for other elementary properties of $C_p(X)$. See this post for a discussion of $C_p(X)$ where $X$ is a separable metric space. See this post about a consequence of normality of $C_p(X)$.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A factorization theorem for products of separable spaces

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $f:X \rightarrow T$ be a continuous function where $T$ is a topological space. In this post, we discuss what it means for the continuous function $f$ to depend on countably many coordinates and then discuss some conditions that we can impose on the product space and on the range space $T$ to ensure that every continuous $f$ defined on the product space will depend on countably many coordinates. This notion of a continuous function depending on countably many coordinates is equivalent to factoring the continuous function into the composition of a projection map and a continuous function defined on a countable subproduct (see Lemma 1 below).

Let’s set some notation about the product space we work with in this post. Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $f:X \rightarrow T$ be continuous. For any $B \subset A$, $\pi_B$ is the natural projection from the full product space $X=\prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in B} X_\alpha$. Standard basic open sets of $X=\prod_{\alpha \in A} X_\alpha$ are of the form $\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is open in $X_\alpha$ and that $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. We use $supp(\prod_{\alpha \in A} O_\alpha)$ to denote the finite set of $\alpha \in A$ where $O_\alpha \ne X_\alpha$.

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Factoring a Continuous Map

The function $f$ is said to depend on countably many coordinates if there exists a countable set $B \subset A$ such that for any $x,y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$. Suppose $f$ is instead defined on a subspace $Y$ of $X$. The function $f$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x,y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.

We have the following lemmas.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $f:X \rightarrow T$ be continuous. Then the following are equivalent.

1. There exists a countable $B \subset A$ such that for any $x,y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.
2. There exists a countable $B \subset A$ such that $f=g \circ \pi_B$ where $g: \prod_{\alpha \in B} X_\alpha \rightarrow T$ is continuous.

Lemma 1a

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $Y$ be a dense subspace of $X$. Let $f:Y \rightarrow T$ be continuous. Then the following are equivalent.

1. There exists a countable $B \subset A$ such that for any $x,y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.
2. There exists a countable $B \subset A$ such that $f=g \circ \pi_B$ where $g: \pi_B(Y) \rightarrow T$ is continuous.

It is straightforward to verify Lemma 1 and Lemma 1a. We use condition 1 to define what it means for a function to be dependent on countably many coordinates. Both lemmas indicate that either condition is a valid definition. These two lemmas also indicate why the notion being discussed can be called a factorization notion.

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When a Continuous Map Can Be Factored

We discuss some conditions that we can place on the product space $X=\prod_{\alpha \in A} X_\alpha$ and on the range space $T$ so that any continuous map depends on countably many coordinates. We prove the following theorem.

Theorem 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space such that each factor $X_\alpha$ is a separable space. Let $T$ be a second countable space (i.e. having a countable base). Then for any dense subspace $Y$ of $X$, any continuous function $f:Y \rightarrow T$ depends on countably many coordinates, i.e., either one of the conditions in Lemma 1a holds.

Before stating the main theorem, we need one more lemma. Let $W \subset X=\prod_{\alpha \in A} X_\alpha$. The set $W$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x \in W$ and for any $y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in W$.

When we try to determine whether a function $f:Y \rightarrow T$, where $Y \subset X$, can be factored, we will need to decide whether a set $W \subset Y$ depends on countably many coordinates. Let $W \subset Y \subset X=\prod_{\alpha \in A} X_\alpha$. The set $W$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x \in W$ and for any $y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in W$. We have the following lemma.

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space with the countable chain condition. Let $Y$ be a dense subspace of $X$.

1. Let $U$ be an open subset of $X$. Then $\overline{U}$ depends on countably many coordinates.
2. Let $W$ be an open subset of $Y$. Then $\overline{W}$ depends on countably many coordinates (closure in $Y$).

Proof of Lemma 2
Proof of Part 1
Let $U \subset X$ be open. Let $\mathcal{B}$ be a collection of pairwise disjoint open subsets of the open set $U \subset X$ such that $\mathcal{B}$ is maximal with this property, i.e., if you throw one more open set into $\mathcal{B}$, it will be no longer pairwise disjoint. Let $V=\bigcup \mathcal{B}$. Since $\mathcal{B}$ is maximal, $\overline{V}=\overline{U}$. Since $X$ has the countable chain condition, $\mathcal{B}$ is countable.

Let $B=\bigcup \left\{supp(O): O \in \mathcal{B} \right\}$. The set $B$ is a countable subset of $A$ since $B$ is the union of countably many finite sets. We have the following claims.

Claim 1
The open set $V$ depends on the coordinates in $B$.

Let $x \in V$ and $y \in X$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show that $y \in V$. Firstly, $x \in O$ for some $O \in \mathcal{B}$. It follows that $x_\alpha=y_\alpha$ for all $\alpha \in supp(O)$. Thus $y \in O \subset V$. This completes the proof of Claim 1.

Claim 2
The set $\overline{V}$ depends on the coordinates in $B$.

Let $x \in \overline{V}$ and $y \in X$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show $y \in \overline{V}$. To this end, let $O=\prod_{\alpha \in A} O_\alpha$ be a standard basic open set with $y \in O$. The goal is to find some $q \in O \cap V$. Define $G=\prod_{\alpha \in A} G_\alpha$ such that $G_\alpha=O_\alpha$ for all $\alpha \in B$ and $G_\alpha=X_\alpha$ for all $\alpha \in A-B$. Then $x \in G$. Since $x \in \overline{V}$, there exists some $p \in V \cap G$. Define $q$ such that $q_\alpha=p_\alpha$ for all $\alpha \in B$ and $q_\alpha=y_\alpha$ for all $\alpha \in A-B$. Since $supp(V) \subset B$, $q \in V$. On the other hand, $q \in O$. This completes the proof of Claim 2.

As noted above, $\overline{V}=\overline{U}$. Thus $\overline{U}$ depends on countably many coordinates, namely the coordinates in the set $B$. This completes the proof of Part 1.

Proof of Part 2
For any $S \subset X$, let $\overline{S}$ denote the closure of $S$ in $Y$. Let $Cl_X(S)$ denote the closure of $S$ in $X$. Let $W \subset Y$ be open. Let $W_1$ be open in $X$ such that $W=W_1 \cap Y$. By Part 1, $Cl_X(W_1)$ depends on countably many coordinates, say the coordinates in the countable set $B \subset A$. This means that for any $x \in Cl_X(W_1)$ and for any $y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in Cl_X(W_1)$. Thus for any $x \in \overline{W}$ and for any $y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in Cl_X(W_1)$. If we have $y \in \overline{W}$, then we are done. So we only need to show that if $y \in Y$ and $y \in Cl_X(W_1)$, then $y \in \overline{W}$. This is why we need to assume $Y$ is dense in $X$.

Let $y \in Y$ and $y \in Cl_X(W_1)$. Let $C$ be an open subset of $Y$ with $y \in C$. There exists an open subset $D$ of $X$ such that $C=D \cap Y$. Then $D \cap Cl_X(W_1) \ne \varnothing$. Note that $D \cap W_1$ is open and $D \cap W_1 \ne \varnothing$. Since $Y$ is dense in $X$, $D \cap W_1$ must contain points of $Y$. These points of $Y$ are also points of $W$. Thus $C$ contains points of $W$. It follows that $y \in \overline{W}$. This concludes the proof of Part 2. $\blacksquare$

Proof of Theorem 1
Let $Y$ be a dense subspace of $X=\prod_{\alpha \in A} X_\alpha$. Let $f:Y \rightarrow T$ be continuous. Let $\mathcal{M}$ be a countable base for the separable metrizable space $T$. By Lemma 2 Part 2, for each $M \in \mathcal{M}$, $\overline{f^{-1}(M)}$ depends on countably many coordinates, say the countable set $B_M$. Let $B=\bigcup_{M \in \mathcal{M}} B_M$.

We claim that $B$ is a countable set of coordinates we need. Let $x,y \in Y$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show that $f(x)=f(y)$. Suppose $f(x) \ne f(y)$. Choose $\left\{M_1,M_2,M_3,\cdots \right\} \subset \mathcal{M}$ such that

• $\left\{f(x) \right\}=\bigcap_{j=1}^\infty M_j=\bigcap_{j=1}^\infty \overline{M_j}$
• $\overline{M_{j+1}} \subset M_j$ for each $j$

This is possible since $T$ is a second countable space. Then $f(y) \notin \overline{M_{k}}$ for some $k$. Furthermore, $y \notin f^{-1}(\overline{M_{k}})$. Since $f$ is continuous, $\overline{f^{-1}(M_{k})} \subset f^{-1}(\overline{M_{k}})$. Therefore, $y \notin \overline{f^{-1}(M_{k})}$. On the other hand, $\overline{f^{-1}(M_{k})}$ depends on the countably many coordinates in $B_{M_k}$. We assume above that $x_\alpha=y_\alpha$ for all $\alpha \in B$. Thus $x_\alpha=y_\alpha$ for all $\alpha \in B_{M_k}$. This means that $y \in \overline{f^{-1}(M_{k})}$, a contradiction. It must be that case that $f(x)=f(y)$. $\blacksquare$

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Another Version

We state another version of Theorem 1 that will be useful in some situations.

Theorem 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space such that each factor $X_\alpha$ is a separable space. Let $T$ be a second countable space. Let $Y$ be a dense subspace of $X$. Let $f:Y \times Y \rightarrow T$ be any continuous function. Then the function $f$ depends on countably many coordinates, which means either one of the following two conditions:

1. There exists a countable set $C \subset A$ such that for any $(x,y),(p,q) \in Y \times Y$, if $x_\alpha=p_\alpha$ and $y_\alpha=q_\alpha$ for all $\alpha \in C$, then $f(x,y)=f(p,q)$.
2. There exists a countable set $C \subset A$ and there exists a continuous $g:\pi_C(Y) \times \pi_C(Y) \rightarrow T$ such that $f=g \circ (\pi_C \times \pi_C)$.

The map $\pi_C \times \pi_C$ is the projection map from $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in C} X_\alpha \times \prod_{\alpha \in B} X_\alpha$ defined by $(\pi_C \times \pi_C)(x,y)=(\pi_C(x),\pi_C(y))$. In Theorem 2, we only need to consider $\pi_C \times \pi_C$ being defined on the subspace $Y \times Y$.

Theorem 2 follows from Theorem 1. It is only a matter of fitting Theorem 2 in the framework of Theorem 1. Note that the product $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ is identical to the product $\prod_{\alpha \in A \cup A^*} X_\alpha$ where $A^*$ is a disjoint copy of the index set $A$. For $(x,y) \in \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$, let $x \times y \in \prod_{\alpha \in A \cup A^*} X_\alpha$ be defined by $(x \times y)_\alpha=x_\alpha$ for all $\alpha \in A$ and $(x \times y)_\alpha=y_\alpha$ for all $\alpha \in A^*$.

With the identification of $(x,y)$ with $x \times y$, we have a setting that fits Theorem 1. The product $\prod_{\alpha \in A \cup A^*} X_\alpha$ is also a product of separable spaces. The set $Y \times Y$ is a dense subspace of the product $\prod_{\alpha \in A \cup A^*} X_\alpha$. In this new setting, we view a point in $Y \times Y$ as $x \times y$. The map $f:Y \times Y \rightarrow T$ is still a continuous map. We can now apply Theorem 1.

Let $B \subset A \cup A^*$ be a countable set such that for all $x \times y,p \times q \in Y \times Y$, if $(x \times y)_\alpha=(p \times q)_\alpha$ for all $\alpha \in B$, then $f(x \times y)=f(p \times q)$. Specifically, if $x_\alpha=p_\alpha$ for all $\alpha \in B \cap A$ and $y_\alpha=q_\alpha$ for all $\alpha \in B \cap A^*$, then $f(x,y)=f(p,q)$.

Choose a countable set $C \subset A$ such that $B \cap A \subset C$ and $B \cap A^* \subset C^*$. Here, $C^*$ is the copy of $C$ in $A^*$. We claim that $C$ is a countable set we need in condition 1 of Theorem 2. Let $(x,y),(p,q) \in Y \times Y$ such that $x_\alpha=p_\alpha$ and $y_\alpha=q_\alpha$ for all $\alpha \in C$. This implies that $x_\alpha=p_\alpha$ for all $\alpha \in B \cap A$ and $y_\alpha=q_\alpha$ for all $\alpha \in B \cap A^*$. Then $f(x,y)=f(p,q)$. Thus condition 1 of Theorem 2 holds. It is also straightforward to verify that Condition 1 and Condition 2 are equivalent.

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Remarks

The notion of factorizing a continuous map defined on a product space is an old topic. Theorem 1 discussed in this post is based on Theorem 4 found in [6]. Theorem 4 found in [6] is to factor continuous maps defined on a product of separable spaces. Theorem 1 in this post is modified to consider continuous maps defined on a dense subspace of a product of separable spaces. This modification will make it more useful. The references listed below represent a small sample of papers or books that have involves theorems of factoring functions defined on products. The work in [3] and [5] have more systematic treatment.

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Reference

1. Brandenburg H., Husek M., On mappings from products into developable spaces, Topology Appl., 26, 229-238, 1987.
2. Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Engelking R., On functions defined on Cartesian products, Fund. Math., 59, 221-231, 1966.
4. Keesling J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
5. Noble N., Ulmer M., Factoring functions on Cartesian products, Trans. Amer. Math. Soc., 163, 329-339, 1972.
6. Ross K. A., Stone A. H., Products of separable spaces, Amer. Math. Monthly, 71, 398-403, 1964.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A lemma dealing with normality in products of separable metric spaces

In this post we prove a lemma that is a great tool for working with product spaces of separable metrizable spaces. As an application of the lemma, we give an alternative proof for showing the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$ where each $X_\alpha$ is a separable and metrizable space. The lemma we discuss here is a tool that can shed some light on normality of dense subspaces of the product space $X$. The lemma is stated in two equivalent forms (Lemma 1 and Lemma 2).

Before stating the lemmas, let’s fix some notations. For any $B \subset A$, the map $\pi_B$ is the natural projection from the full product $X=\prod_{\alpha \in A} X_\alpha$ to the subproduct $\prod_{\alpha \in B} X_\alpha$. The standard basic open sets in the product space $X=\prod_{\alpha \in A} X_\alpha$ are of the form $\prod_{\alpha \in A} O_\alpha$ where $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. We use $supp(\prod_{\alpha \in A} O_\alpha)$ to denote the set of finitely many $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

Given a space $W$, and given $F,G \subset W$, the sets $F$ and $G$ are separated if $F \cap \overline{G}=\varnothing=\overline{F} \cap G$.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. For any sets $H,K \subset Y$, the following two conditions are equivalent:

1. There exist disjoint open subsets $U$ and $V$ of $Y$ such that $H \subset U$ and $K \subset V$.
2. There exists a countable $B \subset A$ such that the sets $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$.

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. Then $Y$ is normal if and only if for each pair of disjoint closed subsets $H$ and $K$ of $Y$, there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$.

If Lemma 1 holds, it is clear that Lemma 2 holds. We prove Lemma 1. The lemmas indicate that to separate disjoint sets in the full product, it suffices to separate in a countable subproduct. In this sense normality in dense subspaces of a product of separable metrizable spaces only depends on countably many coordinates.

This lemma seems to have been around for a long time. We cannot find any reference of this lemma in Engelking’s topology textbook (see [4]). We found three references. One is Corson’s paper (see [3]), in which the lemma is mentioned in relation to the non-normality of $\mathbb{N}^{\omega_1}$ and is attributed to a paper of M. Bockstein in 1948. Another is a paper of Baturov (see [2]), in which the lemma is used to prove a theorem about normality in dense subspace of $M^{\omega_1}$ where $M$ is a separable metric space. In [2] the lemma is attributed to Uspenskii. Another reference is Arkhangelskii’s book on function space (see Lemma I.6.1 on p. 43 in [1]) where the lemma is used in proving some facts about normality in function spaces $C_p(X)$.

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Proof of Lemma 1

$1 \Longrightarrow 2$
Let $U$ and $V$ be disjoint open subsets of $Y$ with $H \subset U$ and $K \subset V$. Let $U_1$ and $V_1$ be open subsets of $X$ such that $U=U_1 \cap Y$ and $V=V_1 \cap Y$. Since $Y$ is dense in $X$, $U_1 \cap V_1=\varnothing$.

Let $\mathcal{U}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $U_1$. Let $\mathcal{V}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $V_1$. These two collections can be obtained using a Zorn lemma argument. The product space $X$ has the countable chain condition since it is a product of separable spaces. So both $\mathcal{U}$ and $\mathcal{V}$ are countable. Let $B$ be the union of finite sets each one of which is a $supp(O)$ where $O \in \mathcal{U} \cup \mathcal{V}$. The set $B$ is countable too.

Let $U^*=\cup \mathcal{U}$ and $V^*=\cup \mathcal{V}$. Note that $U^* \cap V^*=\varnothing$. We have the following observations:

$\pi^{-1}_B(\pi_B(U^*))=U^* \subset U_1$ and $\pi^{-1}_B(\pi_B(V^*))=V^* \subset V_1$

The above observations lead to the following observations:

$\pi^{-1}_B(\pi_B(U^*)) \cap \pi^{-1}_B(\pi_B(V^*)) \subset U_1 \cap V_1=\varnothing$

implying that $\pi_B(U^*) \cap \pi_B(V^*)=\varnothing$. Both $\pi_B(U^*)$ and $\pi_B(V^*)$ are open subsets of $\pi_B(X)$ and are dense in $\pi_B(X)$, respectively.

We claim that $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$. Suppose that $y \in \pi_B(U_1) \cap \pi_B(V_1)$. Then $\pi_B(V_1)$ contains a point of $\pi_B(U^*)$, say $t$. With $t \in \pi_B(U^*)$, $t=\pi_B(q)$ for some $q \in O$ where $O \in \mathcal{U}$. Note that $supp(O) \subset B$. Thus $\pi^{-1}_B(\pi_B(q))=\pi^{-1}_B(t)=O \subset U_1$. On the other hand, $t \in \pi_B(V_1)$ implies that $t=\pi_B(w)$ for some $w \in V_1$. It follows that $w \in U_1 \cap V_1$, a contradiction. Therefore $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$.

We have $\pi_B(H) \subset \pi_B(U) \subset \pi_B(U_1)$ and $\pi_B(K) \subset \pi_B(V) \subset \pi_B(V_1)$. This implies that $\overline{\pi_B(H)} \cap \pi_B(K)=\varnothing$ and $\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$ (closure in $\pi_B(X)$). Then $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$ as well. This concludes the proof for the $1 \Longrightarrow 2$ direction.

$2 \Longrightarrow 1$
Suppose that $B \subset A$ is countable such that $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$. Note that $\pi_B(H) \subset \pi_B(Y)$ and $\pi_B(K) \subset \pi_B(Y)$. Then we have the following:

$\pi_B(H) \cup \pi_B(K) \subset \pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)}) \ \ \ \ \text{closures in } \pi_B(Y)$

Consider $W=\pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)})$. The space $W$ is an open subspace of $\pi_B(Y)$. Furthermore, $\pi_B(Y)$ is a subspace of $\prod_{\alpha \in B} X_\alpha$, which is a separable and metrizable space. Thus the space $W$ is metrizable and hence normal.

For $L \subset W$, let $Cl_W(L)$ denote the closure of $L$ in the space $W$. Note that $Cl_W(\pi_B(H))$ and $Cl_W(\pi_B(K))$ are disjoint and closed sets in $W$. Let $G_H$ and $G_K$ be disjoint open subsets of $W$ such that $Cl_W(\pi_B(H)) \subset G_H$ and $Cl_W(\pi_B(K)) \subset G_K$. Then $\pi^{-1}_B(G_H) \cap Y$ and $\pi^{-1}_B(G_K) \cap Y$ are disjoint open subsets of $Y$ such that $H \subset \pi^{-1}_B(G_H) \cap Y$ and $K \subset \pi^{-1}_B(G_H) \cap Y$. $\blacksquare$

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Remark

The proof of Lemma 1 does not need the full strength of separable metric in each factor of the product space. The above proof only makes two assumptions about the product space: the product space $X=\prod_{\alpha \in A} X_\alpha$ has the countable chain condition (CCC) and that any countable subproduct is normal, i.e., $\prod_{\alpha \in B} X_\alpha$ is normal for any countable $B \subset A$.

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Example

As an application of the above lemma, we give another proof of the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers. See this post for a version of A. H. Stone’s original proof.

Let $\mathbb{N}$ be the set of all nonnegative integers and let $\omega_1$ be the first uncountable ordinal (i.e. the set of all countable ordinals). We provide an alternative proof that $\mathbb{N}^{\omega_1}$ is not normal. In A. H. Stone’s proof, the following disjoint closed sets cannot be separated in $\mathbb{N}^{\omega_1}$:

$H=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 0, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

$K=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 1, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

We can also use Lemma 1 to show that $H$ and $K$ cannot be separated. Note that for each countable $B \subset \omega_1$, the sets $\pi_B(H)$ and $\pi_B(K)$ have non-empty intersection. Hence they cannot be separated in $\pi_B(\mathbb{N}^{\omega_1})$. By Lemma 1, $H$ and $K$ cannot be separated in the full product space $\mathbb{N}^{\omega_1}$.

To see that $\pi_B(H) \cap \pi_B(K) \ne \varnothing$, choose a function $g:\omega_1 \rightarrow \mathbb{N}$ such that $g^{-1}(0) \cap B=\varnothing$. Let $g_B:B \rightarrow \mathbb{N}$ be defined by $g_B(\alpha)=g(\alpha)$ for all $\alpha \in B$. Then $g_B \in \pi_B(H) \cap \pi_B(K)$.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Looking for a closed and discrete subspace of a product space

I had long suspected that there probably is an uncountable closed and discrete subset of the product space of uncountably many copies of the real line. Then I found the statement in the Encyclopedia of General Topology (page 76 in [3]) that “for every infinite cardinal $\mathcal{K}$, the product $D(\mathcal{K})^{2^{\mathcal{K}}}$ includes $D(2^{\mathcal{K}})$ as a closed subspace” where $D(\tau)$ is the discrete space of cardinality of $\tau$. When $\mathcal{K}=\aleph_0$ (the first infinite cardinal), there is a closed and discrete subset of cardinality $c=2^{\aleph_0}$ in the product space $\mathbb{N}^{c}$ (the product space of continuum many copies of a countable discrete space). Despite the fact that this product space is a separable space, a closed and discrete set of cardinality continuum is hiding in the product space $\mathbb{N}^{c}$. What is more amazing is that this result gives us a glimpse into the working of the product topology with uncountably many factors. There are easily defined discrete subspaces of $\mathbb{N}^{c}$. But these discrete subspaces are not closed in the product space, making the result indicated here a remarkable one.

The Encyclopedia of General Topology points to two references [2] and [4]. I could not find these papers online. It turns out that Engelking, the author of [2], included this fact as an exercise in his general topology textbook (see Exercise 3.1.H (a) in [1]). This post presents a proof of this fact based on the hints that are given in [1]. To make the argument easier to follow, the proof uses some of the hints in a slightly different form.

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The Exercise

Let $I=[0,1]$ be the closed unit interval. Let $X=I$ be the unit interval with the discrete topology. Let $\omega$ be the set of all nonnegative integers with the discrete topology. Let $Y=\prod_{t \in I} Y_t$ where each $Y_t=\omega$. We can also denote $Y$ by $\omega^I$. The problem is to show that the discrete space $X$ can be embedded as a closed and discrete subspace of $Y$.

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A Solution

For each $t \in I$, choose a sequence $O_{t,1},O_{t,2},O_{t,3},\cdots$ of open intervals (in the usual topology of $I$) such that

• $t \in O_{t,j}$ for each $j$,
• $\overline{O_{t,j+1}} \subset O_{t,j}$ for each $j$ (the closure is in the usual topology of $I$),
• $\left\{t \right\}=\bigcap \limits_{j=1}^\infty O_{t,j}$.

For any $t \in I-\left\{0,1 \right\}$, we can make $O_{t,j}$ open intervals of the form $(a,b)$. For $t=0$, $O_{0,j}$ have the form $[0,b)$. For $t=1$, $O_{1,j}$ have the form $(a,1]$.

The above sequences of open intervals help define a homeomorphic embedding of the discrete space $X$ into the product $Y$. For each $t \in I$, define the function $f_t:I \rightarrow \omega$ by letting:

$f_t(x) = \begin{cases} 0 & \mbox{if } x=t \\ 1 & \mbox{if } x \in I-O_{t,1} \\ 2 & \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \cdots \\ j & \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \text{etc} \end{cases}$

for each $x \in X$. We now define the embedding $E:X \rightarrow Y=\omega^I$ by letting:

$E(x)=< f_t(x) >_{t \in I}$

for each $x \in X$. For each point $x \in X$, $E(x)$ is the point in the product space such that the $t$ coordinate of $E(x)$ is the value of the function $f_t$ evaluated at $x$. Let $\mathcal{F}=\left\{f_t: t \in I \right\}$.

The mapping $E$ is an evaluation map (called diagonal map in [1]). It is a homeomorphism if the following three conditions are met:

• If each $f_t \in \mathcal{F}$ is continuous, then $E$ is continuous.
• If the family of functions $\mathcal{F}$ separates points in $X$, then $E$ is injective (i.e. a one-to-one function).
• If $\mathcal{F}$ separates points from closed sets in $X$, then the inverse $E^{-1}$ is also continuous.

The first point is easily seen. Note that both the domain and the range of $f_t:X \rightarrow \omega$ have the discrete topology. Thus these functions are continuous. To see the second point, let $p,q \in X$ with $p \ne q$. Note that $f_p(p)=0$ while $f_p(q) \ne 0$.

Since $X$ is discrete, any subset of $X$ is closed. To see the third point, let $C \subset X$ and $p \notin C$. Once again, $f_p(p)=0$ while $f_p(x) \ne 0$ for all $x \in C$. Clearly $f_p(p) \notin \overline{f(C)}$ (closure in the discrete space $\omega$). Thus $E^{-1}$ is also continuous. For more details about why $E$ is an embedding, see the previous post called The Evaluation Map.

Let $W=E(X)$. Since $E$ is a homeomorphism, $W$ is a discrete subspace of the product space $Y$. We only need to show $W$ is closed in $Y$. Let $k \in Y$ such that $k=< k_t >_{t \in I} \ \notin W$. We show that there is an open neighborhood of $k$ that misses the set $W$. There are two cases to consider. One is that $k_r \ne 0$ for all $r \in I$. The other is that $k_r=0$ for some $r \in I$. The first case is more involved.

Case 1
Suppose that $k_r \ne 0$ for all $r \in I$. First define a local base $\mathcal{B}_{k}$ of the point $k$. Let $H \subset I$ be finite and let $G_{H}$ be defined by:

$G_{H}=\left\{b \in Y: \forall \ h \in H, b_h=k_h \right\}$

Let $\mathcal{B}_{k}$ be the set of all possible $G_{H}$. For an arbitrary $G_{H}$, consider $E^{-1}(G_H)$. By definition, $E^{-1}(G_H)=\left\{c \in I: E(c) \in G_H \right\}$. To make the argument below easier to see, let’s further describe $E^{-1}(G_H)$.

\displaystyle \begin{aligned} E^{-1}(G_H)&=\left\{c \in I: E(c) \in G_H \right\} \\&=\left\{c \in I: \forall \ h \in H, f_h(c)=k_h \ne 0 \right\} \\&=\left\{c \in I: \forall \ h \in H, c \in I-O_{h,k_h} \right\} \\&=\bigcap \limits_{h \in H} I-O_{h,k_h} \\&=I-\bigcup \limits_{h \in H} O_{h,k_h} \end{aligned}

The last description above indicates that if $x \in X$ and if $x \in E^{-1}(G_H)$, then $x \notin \bigcup \limits_{h \in H} O_{h,k_h}$. To wrap up Case 1, we would like to produce one particular $H$. Consider the open cover $\left\{O_{t,k_t}: t \in I \right\}$ of $I$. Since $I$ is compact in the usual topology, there is a finite $H \subset I$ such that $I=\bigcup \limits_{h \in H} O_{h,k_h}$. This means that for this particular finite set $H$, $E^{-1}(G_H)=\varnothing$. Putting it in another way, the open neighborhood $B_H$ of $k$ contains no point of $E(x)$. The proof for Case 1 is completed.

Case 2
Suppose that $k_r=0$ for some $r \in I$. Since $k \notin W=E(X)$, in particular $k \ne E(r)=< f_t(r) >_{t \in I}$. So for some $q \in I$, $k_q \ne f_q(r)$. Now define the following open set containing $k$.

$G=\left\{b \in Y: b_r=0 \text{ and } b_q=k_q \right\}$

Note that $E(r) \notin G$ since $k_q \ne f_q(r)$. Furthermore, for each $p \in I-\left\{r \right\}$, $E(p) \notin G$ since $f_r(p) \ne 0$. Thus $G$ is an open neighborhood of $k$ containing no point of $E(X)$. The proof for Case 2 is completed.

We have shown that the image of the discrete space $X$ under the homeomorphism $E$ is closed in the product space $Y=\prod_{t \in I} Y_t=\omega^I$.

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The preceding proof shows that the product space of continuum many copies of $\omega$ contains a closed and discrete subspace of cardinality continuum. This is a remarkable result. At a glance, it is not entirely clear that a closed and discrete set of this large size can be found in the product space in question.

One immediate consequence is that the product space $Y=\prod_{t \in I} Y_t=\omega^I$ is not normal since it is a separable space. By Jones’ lemma, any separable normal space cannot have a closed and discrete subset of cardinality continuum. However, if the goal is only to show non-normality, we only need to show that $\omega^{\omega_1}$ is not normal (a proof is found in this post). Thus the value of the preceding proof is to demonstrate how to produce a closed and discrete subspace of cardinality continuum in the product space in question.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Engelking, R., On the double circumference of Alexandroff, Bull. Acad. Polon. Sci., 16, 629-634, 1968.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Juhasz, I., On closed discrete subspace of product spaces, Bull. Acad. Polon. Sci., 17, 219-223, 1969.

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$\copyright \ 2014 \text{ by Dan Ma}$